ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
6th Edition
ISBN: 9781260406092
Author: HARTWELL, Leland, HOOD, Leroy, Goldberg, Michael
Publisher: Mcgraw-hill Education/stony Brook University
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Chapter 11, Problem 5P
If you examine Fig. 11.5 closely, you will note that in some regions, such as between
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A SNP that can exist as a C or G is associated with a human disease. The genotypic odds ratio of the homozygote CC relative to the homozygote GG is 2.2, which indicates that
a)both CC and GG individuals are equally likely to get the disease because 2.2 is not a significant value.
b)heterozygotes are 2.2 times as likely to get the disease as GG homozygotes.
c) CC homozygotes are 2.2 times as likely to get the disease as GG homozygotes.
d) CC homozygotes are 2.2 times less likely to get the disease as GG homozygotes.
Let’s suppose you make a transposon library of the cellulose-secreting bacterium Komagataeibacter xylinus, with the goal of finding mutants that produce higher than normal amounts of cellulose, which would be useful industrially. However, despite your best efforts you are unable to isolate any transposon mutants that make more cellulose than the wild-type strain.Why might this have failed? List as many reasons as you can think of.
Although DNA transposons are abundant in the genomes of multicellular eukaryotes, class 1 elements usually make up the largest fraction of very large genomes such as those from humans (~2500 Mb), maize (~2500 Mb), and barley (~5000 Mb). Given what you know about class 1 and class 2 elements, what is it about their distinct mechanisms of transposition that would account for this consistent difference in abundance?
Chapter 11 Solutions
ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
Ch. 11 - Choose the phrase from the right column that best...Ch. 11 - Would you characterize the pattern of inheritance...Ch. 11 - Would you be more likely to find single nucleotide...Ch. 11 - A recent estimate of the rate of base...Ch. 11 - If you examine Fig. 11.5 closely, you will note...Ch. 11 - Approximately 50 million SNPs have thus far been...Ch. 11 - Mutations at simple sequence repeat SSR loci occur...Ch. 11 - Humans and gorillas last shared a common ancestor...Ch. 11 - In 2015, an international team of scientists...Ch. 11 - Using PCR, you want to amplify an approximately 1...
Ch. 11 - Prob. 11PCh. 11 - The previous problem raises several interesting...Ch. 11 - You want to make a recombinant DNA in which a PCR...Ch. 11 - You sequence a PCR product amplified from a...Ch. 11 - Prob. 15PCh. 11 - The trinucleotide repeat region of the Huntington...Ch. 11 - Sperm samples were taken from two men just...Ch. 11 - Prob. 18PCh. 11 - a. It is possible to perform DNA fingerprinting...Ch. 11 - On July 17, 1918, Tsar Nicholas II; his wife the...Ch. 11 - The figure that follows shows DNA fingerprint...Ch. 11 - Microarrays were used to determine the genotypes...Ch. 11 - A partial sequence of the wild-type HbA allele is...Ch. 11 - a. In Fig. 11.17b, PCR is performed to amplify...Ch. 11 - The following figure shows a partial microarray...Ch. 11 - Scientists were surprised to discover recently...Ch. 11 - The microarray shown in Problem 25 analyzes...Ch. 11 - The figure that follows shows the pedigree of a...Ch. 11 - One of the difficulties faced by human geneticists...Ch. 11 - Now consider a mating between consanguineous...Ch. 11 - The pedigree shown in Fig. 11.22 was crucial to...Ch. 11 - You have identified a SNP marker that in one large...Ch. 11 - The pedigrees indicated here were obtained with...Ch. 11 - Approximately 3 of the population carries a mutant...Ch. 11 - The drug ivacaftor has recently been developed to...Ch. 11 - In the high-throughput DNA sequencing protocol...Ch. 11 - A researcher sequences the whole exome of a...Ch. 11 - As explained in the text, the cause of many...Ch. 11 - Figure 11.26 portrayed the analysis of Miller...Ch. 11 - A research paper published in the summer of 2012...Ch. 11 - Table 11.2 and Fig. 11.27 together portray the...Ch. 11 - The human RefSeq of the entire first exon of a...Ch. 11 - Mutations in the HPRT1 gene in humans result in at...Ch. 11 - Prob. 44P
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- You learned in Problem 21 in Chapter 7 that theneurodegenerative disease ALS can be caused by expansion of a hexanucleotide repeat region (5′-GGGGCC-3′)outside of the open reading frame (but within the firstintron) of the gene called C9ORF72. While a normalC9ORF72 allele has 2–23 copies of the hexanucleotiderepeat unit, dominant disease-causing alleles have hundreds or even thousands of copies. Researchers observed that the first intron of theC9ORF72 disease allele is transcribed not only fromthe normal template strand of DNA, but also from thenontemplate strand. Even more unusual, both types ofrepeat-region transcripts are translated in all six readingframes in an AUG-independent manner—a processcalled repeat-associated non-ATG translation, or RANtranslation. These discoveries led to the hypothesisthat the proteins made from the repeats mightcontribute to ALS.a. What polypeptides are made from the repeat-regiontranscripts?b. According to the RAN translation hypothesis, whyare…arrow_forwardThe DNA-binding domain of each CREB protein subunit recognizes the sequence 5′–TGACGTCA–3′. Due to random chance, how often would you expect this sequence to occur in the human genome, which contains approximately 3 billion base pairs? Actually, only a few doze genes are activated by the CREB protein. Does the value of a few dozen agree with the number of random occurrences expected in the human genome? If the number of random occurrences of the sequence in the human genome is much higher than a few dozen, provide at least one explanation why the CREB protein is not activating more than a few dozen gene Actually, only a few doze genes are activated by the CREB protein. Does the value of a few dozen agree with the number of random occurrences expected in the human genome? If the number of random occurrences of the sequence in the human genome is much higher than a few dozen, provide at least one explanation why the CREB protein is not activating more than a few dozen genearrow_forwardIn addition to the standard base-paired helical structures, DNA can form X-shaped hairpin structures called cruciforms in which most bases are involved in Watson–Crick pairs. Such structures tend to occur at sequences with inverted repeats. Draw the cruciform structure formed by the DNA sequence TCAAGTCCACGGTGGACTTGC.arrow_forward
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